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%I #35 Jan 21 2022 03:43:50
%S 1,5,13,14,15,17,27,29,37,39,41,46,51,53,55,61,73,87,89,95,97,101,109,
%T 111,113,123,124,137,142,143,149,157,159,173,181,183,186,187,193,197,
%U 206,215,219,229,231,233,241,247,257,267,269,277,279,281,291,293,295,302,303,313,317,319,323
%N Numbers k such that the number of divisors of k is equal to the number of power-of-two-divisors of the sum of divisors of k.
%C Numbers k such that A000005(k) = A286357(k). - _Antti Karttunen_, Aug 12 2019
%C Perfect powers: 1, 27, 1331, 6859, 79507, 205379, 300763, 571787, 1225043, 2248091, 2685619, 4330747, 5735339, 9393931, ...
%F A000005(a(n)) = A001511(A000203(a(n))).
%e 14 is in this sequence because it has 4 divisors (1, 2, 7, 14) and 1 + 2 + 7 + 14 = 24 also has 4 power-of-two-divisors 1, 2, 4, 8.
%t Select[Range[323], DivisorSigma[0, #] == 1 + IntegerExponent[ DivisorSigma[1, #], 2] &] (* _Giovanni Resta_, Sep 19 2019 *)
%o (Magma) [n: n in [1..300] | NumberOfDivisors(n) eq Valuation(2*SumOfDivisors(n),2)];
%o (PARI) isA309621(n) = (numdiv(n)==(1+valuation(sigma(n),2))); \\ _Antti Karttunen_, Aug 12 2019
%Y Supersequence of A002144.
%Y Cf. A000005, A000203, A001511, A001597, A286357, A309530.
%K nonn,easy
%O 1,2
%A _Juri-Stepan Gerasimov_, Aug 10 2019